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feat: add solutions to lc problem: No.0956
No.0956.Tallest Billboard
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Diff for: ‎solution/0900-0999/0956.Tallest Billboard/README.md

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@@ -50,22 +50,169 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划**
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我们定义 $f[i][j]$ 表示前 $i$ 根钢筋，两边高度差为 $j$ 时的最大共同高度。初始时 $f[0][0]=0$，其余 $f[i][j]=-\infty$。我们求出所有 $rods[i]$ 的和，记为 $s$，那么 $j$ 的取值范围为 $[0,..s]$。
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对于第 $i$ 根钢筋，我们可以不选择它，此时 $f[i][j]=f[i-1][j]$；也可以选择它，此时有三种情况：
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1. 放置在原本高度较高的一边，即满足 $j \geq rods[i-1]$，此时 $f[i][j] = max(f[i][j], f[i-1][j-rods[i-1]])$；
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1. 放置在原本高度较低的一遍，如果满足 $j + rods[i-1] \leq s$，此时 $f[i][j] = max(f[i][j], f[i-1][j+rods[i-1]] + rods[i-1])$；如果满足 $j \lt rods[i-1]$，此时 $f[i][j] = max(f[i][j], f[i-1][rods[i-1]-j] + rods[i-1]-j)$。
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综上，我们可以得到状态转移方程：
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$$
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\begin{aligned}
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f[i][j] &= f[i-1][j] \\
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f[i][j] &= max(f[i][j], f[i-1][j-rods[i-1]]) & \text{if } j \geq rods[i-1] \\
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f[i][j] &= max(f[i][j], f[i-1][j+rods[i-1]] + rods[i-1]) & \text{if } j + rods[i-1] \leq s \\
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f[i][j] &= max(f[i][j], f[i-1][rods[i-1]-j] + rods[i-1]-j) & \text{if } j \lt rods[i-1]
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\end{aligned}
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$$
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最终答案即为 $f[n][0]$。
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时间复杂度 $O(n \times S)$，空间复杂度 $O(n \times S)$。其中 $n$ 和 $S$ 分别为 $rods$ 的长度和 $rods$ 中所有元素的和。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def tallestBillboard(self, rods: List[int]) -> int:
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n = len(rods)
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s = sum(rods)
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f = [[-inf] * (s + 1) for _ in range(n + 1)]
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f[0][0] = 0
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t = 0
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for i, x in enumerate(rods, 1):
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t += x
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for j in range(t + 1):
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f[i][j] = f[i - 1][j]
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if j >= x:
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f[i][j] = max(f[i][j], f[i - 1][j - x])
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if j + x <= t:
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f[i][j] = max(f[i][j], f[i - 1][j + x] + x)
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if j < x:
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f[i][j] = max(f[i][j], f[i - 1][x - j] + x - j)
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return f[n][0]
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int tallestBillboard(int[] rods) {
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int n = rods.length;
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int s = 0;
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for (int x : rods) {
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s += x;
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}
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int[][] f = new int[n + 1][s + 1];
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for (var e : f) {
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Arrays.fill(e, -(1 << 30));
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}
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f[0][0] = 0;
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for (int i = 1, t = 0; i <= n; ++i) {
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int x = rods[i - 1];
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t += x;
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for (int j = 0; j <= t; ++j) {
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f[i][j] = f[i - 1][j];
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if (j >= x) {
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f[i][j] = Math.max(f[i][j], f[i - 1][j - x]);
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}
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if (j + x <= t) {
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f[i][j] = Math.max(f[i][j], f[i - 1][j + x] + x);
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}
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if (j < x) {
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f[i][j] = Math.max(f[i][j], f[i - 1][x - j] + x - j);
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}
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}
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}
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return f[n][0];
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int tallestBillboard(vector<int>& rods) {
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int n = rods.size();
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int s = accumulate(rods.begin(), rods.end(), 0);
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int f[n + 1][s + 1];
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memset(f, -0x3f, sizeof(f));
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f[0][0] = 0;
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for (int i = 1, t = 0; i <= n; ++i) {
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int x = rods[i - 1];
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t += x;
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for (int j = 0; j <= t; ++j) {
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f[i][j] = f[i - 1][j];
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if (j >= x) {
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f[i][j] = max(f[i][j], f[i - 1][j - x]);
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}
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if (j + x <= t) {
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f[i][j] = max(f[i][j], f[i - 1][j + x] + x);
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}
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if (j < x) {
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f[i][j] = max(f[i][j], f[i - 1][x - j] + x - j);
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}
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}
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}
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return f[n][0];
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}
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};
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```
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### **Go**
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```go
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func tallestBillboard(rods []int) int {
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n := len(rods)
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s := 0
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for _, x := range rods {
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s += x
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}
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f := make([][]int, n+1)
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for i := range f {
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f[i] = make([]int, s+1)
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for j := range f[i] {
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f[i][j] = -(1 << 30)
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}
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}
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f[0][0] = 0
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for i, t := 1, 0; i <= n; i++ {
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x := rods[i-1]
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t += x
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for j := 0; j <= t; j++ {
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f[i][j] = f[i-1][j]
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if j >= x {
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f[i][j] = max(f[i][j], f[i-1][j-x])
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}
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if j+x <= t {
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f[i][j] = max(f[i][j], f[i-1][j+x]+x)
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}
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if j < x {
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f[i][j] = max(f[i][j], f[i-1][x-j]+x-j)
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}
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}
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}
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return f[n][0]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

Diff for: ‎solution/0900-0999/0956.Tallest Billboard/README_EN.md

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@@ -51,13 +51,136 @@
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### **Python3**
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```python
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class Solution:
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def tallestBillboard(self, rods: List[int]) -> int:
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n = len(rods)
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s = sum(rods)
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f = [[-inf] * (s + 1) for _ in range(n + 1)]
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f[0][0] = 0
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t = 0
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for i, x in enumerate(rods, 1):
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t += x
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for j in range(t + 1):
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f[i][j] = f[i - 1][j]
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if j >= x:
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f[i][j] = max(f[i][j], f[i - 1][j - x])
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if j + x <= t:
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f[i][j] = max(f[i][j], f[i - 1][j + x] + x)
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if j < x:
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f[i][j] = max(f[i][j], f[i - 1][x - j] + x - j)
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return f[n][0]
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```
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### **Java**
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```java
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class Solution {
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public int tallestBillboard(int[] rods) {
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int n = rods.length;
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int s = 0;
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for (int x : rods) {
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s += x;
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}
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int[][] f = new int[n + 1][s + 1];
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for (var e : f) {
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Arrays.fill(e, -(1 << 30));
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}
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f[0][0] = 0;
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for (int i = 1, t = 0; i <= n; ++i) {
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int x = rods[i - 1];
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t += x;
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for (int j = 0; j <= t; ++j) {
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f[i][j] = f[i - 1][j];
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if (j >= x) {
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f[i][j] = Math.max(f[i][j], f[i - 1][j - x]);
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}
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if (j + x <= t) {
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f[i][j] = Math.max(f[i][j], f[i - 1][j + x] + x);
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}
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if (j < x) {
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f[i][j] = Math.max(f[i][j], f[i - 1][x - j] + x - j);
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}
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}
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}
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return f[n][0];
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int tallestBillboard(vector<int>& rods) {
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int n = rods.size();
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int s = accumulate(rods.begin(), rods.end(), 0);
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int f[n + 1][s + 1];
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memset(f, -0x3f, sizeof(f));
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f[0][0] = 0;
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for (int i = 1, t = 0; i <= n; ++i) {
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int x = rods[i - 1];
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t += x;
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for (int j = 0; j <= t; ++j) {
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f[i][j] = f[i - 1][j];
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if (j >= x) {
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f[i][j] = max(f[i][j], f[i - 1][j - x]);
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}
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if (j + x <= t) {
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f[i][j] = max(f[i][j], f[i - 1][j + x] + x);
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}
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if (j < x) {
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f[i][j] = max(f[i][j], f[i - 1][x - j] + x - j);
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}
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}
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}
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return f[n][0];
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}
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};
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```
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### **Go**
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```go
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func tallestBillboard(rods []int) int {
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n := len(rods)
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s := 0
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for _, x := range rods {
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s += x
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}
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f := make([][]int, n+1)
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for i := range f {
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f[i] = make([]int, s+1)
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for j := range f[i] {
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f[i][j] = -(1 << 30)
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}
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}
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f[0][0] = 0
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for i, t := 1, 0; i <= n; i++ {
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x := rods[i-1]
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t += x
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for j := 0; j <= t; j++ {
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f[i][j] = f[i-1][j]
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if j >= x {
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f[i][j] = max(f[i][j], f[i-1][j-x])
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}
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if j+x <= t {
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f[i][j] = max(f[i][j], f[i-1][j+x]+x)
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}
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if j < x {
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f[i][j] = max(f[i][j], f[i-1][x-j]+x-j)
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}
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}
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}
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return f[n][0]
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

Diff for: ‎solution/0900-0999/0956.Tallest Billboard/Solution.cpp

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@@ -0,0 +1,27 @@
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class Solution {
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public:
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int tallestBillboard(vector<int>& rods) {
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int n = rods.size();
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int s = accumulate(rods.begin(), rods.end(), 0);
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int f[n + 1][s + 1];
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memset(f, -0x3f, sizeof(f));
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f[0][0] = 0;
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for (int i = 1, t = 0; i <= n; ++i) {
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int x = rods[i - 1];
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t += x;
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for (int j = 0; j <= t; ++j) {
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f[i][j] = f[i - 1][j];
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if (j >= x) {
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f[i][j] = max(f[i][j], f[i - 1][j - x]);
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}
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if (j + x <= t) {
18+
f[i][j] = max(f[i][j], f[i - 1][j + x] + x);
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}
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if (j < x) {
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f[i][j] = max(f[i][j], f[i - 1][x - j] + x - j);
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}
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}
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}
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return f[n][0];
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}
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};

Diff for: ‎solution/0900-0999/0956.Tallest Billboard/Solution.go

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func tallestBillboard(rods []int) int {
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n := len(rods)
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s := 0
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for _, x := range rods {
5+
s += x
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}
7+
f := make([][]int, n+1)
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for i := range f {
9+
f[i] = make([]int, s+1)
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for j := range f[i] {
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f[i][j] = -(1 << 30)
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}
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}
14+
f[0][0] = 0
15+
for i, t := 1, 0; i <= n; i++ {
16+
x := rods[i-1]
17+
t += x
18+
for j := 0; j <= t; j++ {
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f[i][j] = f[i-1][j]
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if j >= x {
21+
f[i][j] = max(f[i][j], f[i-1][j-x])
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}
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if j+x <= t {
24+
f[i][j] = max(f[i][j], f[i-1][j+x]+x)
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}
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if j < x {
27+
f[i][j] = max(f[i][j], f[i-1][x-j]+x-j)
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}
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}
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}
31+
return f[n][0]
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}
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34+
func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}

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